Remainder Number Theory Problem

by | Mar 14, 2020 | GMAT

Fatma Xhafa

Fatma Xhafa

Fatma is our Director of Operations and has been with Apex since 2020. With over four years of experience in the test prep industry, she combines her background in computer science and business with deep industry expertise to manage day-to-day operations and lead the Apex team.

Today we’re going to be looking at the Remainder Number Theory problem:

If 23 is divided by some integer n, the remainder is 3.

I. Is n even?
II. Is n a multiple of 5?
III. Is n a factor of 120?

(A) I and II only
(B) III only
(C) I, II, and III
(D) II and III only
(E) II only

Our big question is that originally we’re given this unknown number N and we know we just have a remainder 3. So the problem is presenting us information in a way that we’re not used to seeing and what we need to do is work backwards from this to drive the core insights.

Sorting Through the Information

So if we have a remainder of 3 on 23 this means that the chunk that isn’t remainder is 20. So what can our n be in those cases that will allow us to divide out by 20 and leave this remainder 3?

Well, first we know that n has to be greater than 3 because in order to have a remainder the amount we’re dividing by has to be something greater. The moment the remainder equalizes the thing we’re dividing by of course we get one more tick in the dividing by box and the remainder goes back down to zero.

Solving

So with 23 and a remainder of 3, our key number to look at is 20. Our factors of 20, that is the things that divide evenly into 20, are 1, 2, 4, 5, 10, and 20. Of course, 1 and 2 are below 3 and so they’re not contenders. So we end up with n being 4, 5, 10, or 20.

Check Against the Statements

So for number 1: Is N even? If N can be 4 but can also be 5 then we’re not assured that it’s even. Notice the data sufficiency problem type embedded here. So N is not necessarily even.

Is N a multiple of 5? Once again N is not because N could be 4 or 5. Finally, is N a factor of 20? And in this case, it is because 4, 5, 10, and 20 as we just said are all factors of that 20 that we’re looking for. So our answer here is 3 alone, answer choice A.

More Practice

Now here’s a more challenging problem at the same form, see if you can do it and we’re going to come back and in the next video talk about the solution and give you another problem.

So if 67 is divided by some integer N the remainder is 7. Our three things to look at are whether:

    • N is even?
    • If N is a multiple of 10?
    • Or N is a factor of 120?

So give this one a try and see if you can use the principles from the easier problem on this more challenging one to make sure that you actually understand what’s going on. If not, re-watch this video and see if a review might allow you to answer this question.

If you enjoyed using this video for practice, try this one next: Wedding Guest Problem.